Ch. 24 Questions

Number of fingers in humans. Extra (more than five) fingers are caused by the presence of an autosomal dominant allele.

A discontinuous characteristic because there are only a few distinct phenotypes determined by alleles at a single locus.

Number of toes in guinea pigs, which is influenced by genes at many loci.

A quantitative characteristic because it is determined by many loci. The number of toes is an example of a meristic characteristic.

How does a quantitative characteristic differ from a discontinuous characteristic?

Discontinuous characteristics have only a few distinct (discrete) phenotypes. In contrast, a quantitative characteristic shows a continuous variation in phenotype.

A strawberry farmer determines that the average weight of individual strawberries produced by plants in his garden is 2 g. He selects the 10 plants that produce the largest strawberries; the average weight of strawberries among these selected plants is 6 g. He interbreeds these selected plants. The progeny of these selected plants produce strawberries that weigh 5 g. If the farmer were to select plants that produce an average strawberry weight of 4 grams, what would be the predicted weight of strawberries produced by the progeny of these selected plants?

Here we can use the equation R = h × S. R, the response to selection, is the difference between the mean of the starting population and the mean of the progeny of the selected parents. In this case, R = 5 g - 2 g = 3 g. S, the selection differential, is the difference between the mean of the starting population and the mean of the selected parents; in this case S = 6 g - 2 g = 4 g. Substituting in the equation, we get 3 g = h2(4 g); h2 = 0.75. If the selected plants averaged 4 g, then S would be 2 g and R = 0.75(2 g) = 1.5 g. Therefore, the predicted average weight of strawberries from the progeny plants would be 2 g + 1.5 g = 3.5 g.

A genetics researcher determines that the broad-sense heritability of height among Southwestern University undergraduate students is 0.90. Which of the following conclusions would be reasonable? Explain your answer. a. Since Sally is a Southwestern University undergraduate student, 10% of her height is determined by nongenetic factors. b. Ninety percent of variation in height among all undergraduate students in the United States is due to genetic differences. c. Ninety percent of the height of Southwestern University undergraduate students is determined by genes. d. Ten percent of the variation in height of Southwestern University undergraduate students is determined by variation in nongenetic factors. e. Because the heritability of height among Southwestern University students is so high, any change in the students' environment will have minimal impact on their height.

Heritability is the proportion of total phenotypic variance that is due to genetic variance, and applies only to the particular population. Thus, the only reasonable conclusion is (d). Statement (a) is not justified because the heritability value does not apply to absolute height nor to an individual, but to the variance in height among Southwestern undergraduates. Statement (b) is not justified because the heritability has been determined only for Southwestern University students; students at other universities, with different ethnic backgrounds and from different regions of the country may have different heritability for height. Statement (c) is again not justified because the heritability refers to the variance in height rather than absolute height. Statement (e) is not justified because the heritability has been determined for the range of variation in nongenetic factors experienced by the population under study; environmental variation outside this range (such as severe malnutrition) may have profound effects on height.

The narrow-sense heritability of ear length in Reno rabbits is 0.4. The phenotypic variance (VP) is 0.8, and the environmental variance (VE) is 0.2. What is the additive genetic variance (VA) for ear length in these rabbits?

Narrow-sense heritability = VA/VP = 0.4 Given that VP = 0.8, VA = 0.4(0.8) = 0.32

Briefly explain why the relation between genotype and phenotype is frequently complex for quantitative characteristics.

Quantitative characteristics are polygenic, so many genotypes are possible. Moreover, most quantitative characteristics are also influenced by environmental factors. Therefore, the phenotype is determined by complex interactions of many possible genotypes and environmental factors.

Assume that three loci, each with two alleles (A and a, B and b, C and c), determine the differences in height between two homozygous strains of a plant. These genes are additive and equal in their effects on plant height. One strain (aa bb cc) is 10 cm in height. The other strain (AA BB CC) is 22 cm in height. The two strains are crossed, and the resulting F1 are interbred to produce F2 progeny. Give the phenotypes and the expected proportions of the F2 progeny.

The AABBCC strain is 12 cm taller than the aabbcc strain. We therefore calculate that each dominant allele adds 2 cm of height above the baseline 10 cm of the all-recessive strain. The F1, with genotype AaBbCc, therefore will be 10 + 6 = 16 cm tall. The seven different possible phenotypes with respect to plant height and the expected frequencies in the F2 are listed in the following table: # of dominant alleles; height; proportion of F2 progeny: 6 22 1/64 5 20 6/64 4 18 15/64 3 16 20/64 2 14 15/64 1 12 6/64 0 10 1/64 The proportions can be determined by counting the numbers of boxes with one dominant allele, two dominant alleles, and so on from an 8 × 8 Punnett square.

How do broad-sense and narrow-sense heritabilities differ?

The broad-sense heritability is the portion of phenotypic variance that is due to all types of genetic variance, including additive, dominance, and genic interaction variances. The narrow-sense heritability is only that portion of the phenotypic variance that is due to additive genetic variance.

Flower color in the varieties of pea plants studied by Mendel is controlled by alleles at a single locus. A group of peas homozygous for purple flowers is grown. Careful study of the plants reveals that all their flowers are purple, but there is some variability in the intensity of the purple color. What would the estimated heritability be for the variation in flower color? Explain your answer.

The plants are homozygous for the single color locus; therefore, there is no genetic variance: VG = 0. Because heritability is VG/VP, if VG is zero, then heritability is zero.

Kernel color in a strain of wheat, in which two codominant alleles segregating at a single locus determine the color. Thus, there are three phenotypes present in this strain: white, light red, and medium red.

This is a discontinuous characteristic because only a few distinct phenotypes are present and it is determined by alleles at a single locus.

Body weight in a family of Labrador retrievers. An autosomal recessive allele that causes dwarfism is present in this family. Two phenotypes are recognized: dwarf (less than 13 kg) and normal (greater than 23 kg).

This is a discontinuous characteristic because there are only two phenotypes (dwarf and normal) and a single locus determines the characteristic.

Presence or absence of leprosy. Susceptibility to leprosy is determined by multiple genes and numerous environmental factors.

This is a quantitative characteristic because susceptibility is a continuous trait that is determined by multiple genes and environmental factors. It is an example of a quantitative phenotype with a threshold effect.

List all the components that contribute to the phenotypic variance and define each component.

VG - Component of variance due to variation in genotype VA - Component of variance due to additive genetic variance VD - Component of variance due to dominance genetic variance VI - Component of variance due to genic interaction variance VE - Component of variance due to environmental differences VGE - Component of variance due to interaction between genes and environment

A rancher determines that the average amount of wool produced by a sheep in her flock is 22 kg per year. In an attempt to increase the wool production of her flock, the rancher picks five male and five female sheep with the greatest wool production; the average amount of wool produced per sheep by those selected is 30 kg. She interbreeds these selected sheep and finds that the average wool production among the progeny of the selected sheep is 28 kg. What is the narrow-sense heritability for wool production among the sheep in the rancher's flock?

We use the equation R = h × S. The value of R is given by the difference in the average wool production of the progeny of the selected sheep compared to the rest of the flock: 28 kg - 22 kg = 6 kg. The value of S is the difference between the selected sheep and the flock: 30 kg - 22 kg = 8 kg. Then, h2 = R/S = 6/8 = 0.75.

Assume that plant weight is determined by a pair of alleles at each of two independently assorting loci (A and a, B and b) that are additive in their effects. Further assume that each allele represented by an uppercase letter contributes 4 g to weight and that each allele represented by a lowercase letter contributes 1 g to weight. a. If a plant with genotype AA BB is crossed with a plant with genotype aa bb, what weights are expected in the F1 progeny? b. What is the distribution of weight expected in the F2 progeny?

a. All the F1 progeny will have genotype Aa Bb, so they should all have 4 + 1 + 4 + 1 = 10 grams of weight. b. We can group the 16 expected genotypes by the number of uppercase and lowercase alleles: 4 uppercase: AA BB = 1/16 with 16 grams 3 uppercase: 2 Aa BB, 2 AA Bb = 4/16 with 13 grams 2 uppercase: 4 Aa Bb, aa BB, AA bb = 6/16 with 10 grams 1 uppercase: 2 Aa bb, 2 aa Bb = 4/16 with 7 grams 0 uppercase: aa bb = 1/16 with 4 grams

Phenotypic variation in tail length of mice has the following components: 𝑥2 Additive genetic variance (VA) = 0.5 Dominance genetic variance (VD) =0.3 Genic interaction variance (VI) =0.1 Environmental variance (VE) = 0.4 Genetic-environmental interaction variance (VGE) = 0.0 a. What is the narrow-sense heritability of tail length? Solution: b. What is the broad-sense heritability of tail length?

a. Narrow-sense heritability is VA/VP = 0.5/1.3 = 0.38. b. Broad-sense heritability is VG/VP = (VA + VD + VI)/VP = 0.9/1.3 = 0.69. c.

A characteristic has a narrow-sense heritability of 0.6. a. If the dominance variance (VD) increases and all other variance components remain the same, what will happen to the narrow-sense heritability? Will it increase, decrease, or remain the same? Explain. b. What will happen to the broad-sense heritability? Explain. c. If the environmental variance (VE) increases and all other variance components remain the same, what will happen to the narrow-sense heritability? Explain. d. What will happen to the broad-sense heritability?

a. The narrow-sense heritability will decrease. Narrow-sense heritability is VA/VP. Increasing the VD will increase the total phenotypic variance VP. If VA remains unchanged, then the proportion VA/VP will become smaller. b. The broad-sense heritability VG/VP will increase. VG is the sum of VA + VD + VI. VP is the sum of VA + VD + VI + VE. Increasing the numerator and denominator of the fraction by the same arithmetic increment will result in a larger fraction, if the fraction is smaller than 1, as must be the case for VG/VP. c. The narrow sense heritability VA/VP will decrease because the total phenotypic variance VP will increase if VE increases. d. The broad-sense heritability VG/VP will decrease because VP = VG + VE will increase.

Joe is breeding cockroaches in his dorm room. He finds that the average wing length in his population of cockroaches is 4 cm. He chooses six cockroaches that have the largest wings; the average wing length among these selected cockroaches is 10 cm. Joe interbreeds these selected cockroaches. From earlier studies, he knows that the narrow-sense heritability for wing length in his population of cockroaches is 0.6. a. Calculate the selection differential and expected response to selection for wing length in these cockroaches. b. What should be the average wing length of the progeny of the selected cockroaches?

a. Usetheequation:R=h ×S,whereSistheselectiondifferential.Inthiscase,S = 10 cm - 4 cm = 6 cm, and we are given that the narrow-sense heritability h2 is 0.6. Therefore, the response to selection R = 0.6(6 cm) = 3.6 cm. b. The average wing length of the progeny should be the mean wing length of the population plus R: 4 cm + 3.6 cm = 7.6 cm.

Assume that human ear length is influenced by multiple genetic and environmental factors. Suppose you measured ear length on three groups of people, in which group A consists of five unrelated persons, group B consists of five siblings, and group C consists of five first cousins. a. With the assumption that the environment for each group is similar, which group should have the highest phenotypic variance? Explain why. b. Is it realistic to assume that the environmental variance for each group is similar? Explain your answer.

a. Group A, because unrelated individuals have the greatest genetic variance. b. No. Siblings from the same family and who are raised in the same house should have smaller environmental variance than group A of unrelated individuals.